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\newtheorem{definition}{Definition}[section]%定义
\newtheorem{theorem}{Theorem}[section]%定理
\newtheorem{axiom}{Axiom}[section]%公理
\newtheorem{lemma}{Lemma}[section]%引理
\newtheorem{proposition}{Proposition}[section]%命题
\newtheorem{corollary}{Corollary}[section]%推论
\newtheorem{remark}{Remark}[section]%注


\title{\heiti\zihao{2} 复变函数-第2章}
\author{中书君}
\date{\today}

\begin{document}
\maketitle
\section{函数$w=\dfrac{1}{z}$把平面上的下列曲线映射成$w$平面上的什么曲线(其中$z=x+iy,w=u+iv$)?}
\textbf{解:}\quad
$w=\dfrac{1}{z}=\dfrac{1}{x+yi}=\dfrac{x-yi}{x^2+y^2}=u+vi$

(1)$x^2+y^2=4$

令$x=2\cos\theta,y=2\sin\theta$,则$w=\dfrac{2\cos\theta}{4}-\dfrac{2i\sin\theta}{4}$
是以$\dfrac{1}{2}$为半径,$0$为原点的圆.($u^2+v^2=\dfrac{1}{4}$)

(2)$y=x$

$w=\dfrac{x}{2x^2}-\dfrac{x}{2x^2}i=\dfrac{1}{2x}-\dfrac{1}{2x}i(x\neq 0)$.其为$u=-v$这条直线除去$0$点.

(3)$y=0$;
$w = \dfrac{1}{x}$,为$y=0$除去$(0,0)$

(4)$x=1$

$w = \dfrac{1}{1+y^2} -i\dfrac{y}{1+y^2}$

$$
	\begin{aligned}
		4\left(u - \dfrac{1}{2}\right)^2 + 4v^2 & = 1            \\
		\left(u - \dfrac{1}{2}\right)^2 + v^2   & = \dfrac{1}{4}
	\end{aligned}
$$

其为以$\dfrac{1}{2}$为圆心,半径为$\dfrac{1}{2}$的圆.
\section{设函数
  $$
	  f(z)=\left\{\begin{array}{ll}
		  \dfrac{x y}{x^{2}+y^{2}}, & z \neq 0 \\
		  0,                        & z=0
	  \end{array}\right.
  $$
  证明： $f(z)$ 在原点不连续.}
\begin{proof}
	令$y=kx$,即$z=x+kxi$,则$\lim\limits_{z\rightarrow 0}f(z)=\lim\limits_{(x,y)\rightarrow (0,0)}\dfrac{xy}{x^2+y^2}=\lim\limits_{x\rightarrow 0}\dfrac{kx^2}{(1+k^2)x^2}=\dfrac{k}{1+k^2}$,与$k$有关,从而$(x,y)$沿着不同的方式趋近$(0,0)$时的极限不相等,所以$f(z)$在$(0,0)$不连续.
\end{proof}

\section{设函数 $f(z)$ 在 $z_{0}$ 连续, 且 $f\left(z_{0}\right) \neq 0$, 那么可以找到 $z_{0}$ 的一个邻域,在这个邻域内 $f(z) \neq 0$.}
\begin{proof}
	令$f(z_0)=a$,则由于$f(z)$连续,从而$\lim\limits_{z \rightarrow z_0} = f(z_0) = a$,从而对于$\delta = \dfrac{a}{2}\exists \varepsilon > 0,s.t.\forall z, |z-z_0|<\varepsilon, |f(z) - a |< \dfrac{a}{2}$.从而可以找到 $z_{0}$ 的一个邻域,在这个邻域内 $f(z) \neq 0$.
\end{proof}

\section{试证明连续函数$f(z)$的模也是连续的.}
\begin{proof}
	连续函数$f(z) = f(u(x,y)+iv(x,y))$,其中$u,v$是关于$x,y$的连续函数.从而$|f(z)| = \sqrt{u^2+v^2}$是连续函数(分析学中,我们已经知道连续函数的初等复合函数都是连续函数).
\end{proof}


\section{试证$\arg z$在原点与负实轴上不连续}
\begin{proof}
	(1)令$z=a+kai,k>0$,则$\lim\limits_{z\rightarrow 0}\arg z = \arctan k$,与$k$有关,从而在$0$点处不连续.

	(2)令$z = a + xi$,当$x>0$时,$\lim\limits_{z\rightarrow a}\arg z = \pi$,而当$x<0$时,$\lim\limits_{z\rightarrow a}\arg z = -\pi$.二者极限不相等,从而在负实轴上处处不连续.
\end{proof}

\section{下列函数在何处有导数?并求出其导数}
 (1)$(z-1)^n$
$$
	\begin{aligned}
		\dfrac{f(z + \Delta z) - f(z)}{\Delta z} & =\dfrac{(z + \Delta z - 1)^n-(z-1)^n}{\Delta z}       \\
		                                         & =\dfrac{n\Delta z(z-1)^{n-1} + o(\Delta z)}{\Delta z} \\
		                                         & =n(z-1)^n
	\end{aligned}
$$
在复平面上任意一点可导.

(2)$\dfrac{1}{z^2-1}$
当$\Delta z \rightarrow 0$时:
$$
	\begin{aligned}
		\dfrac{f(z+\Delta z) - f(z)}{\Delta z} & =\dfrac{\dfrac{1}{(z+\Delta z)^2-1}-\dfrac{1}{z^2-1}}{\Delta z}                 \\
		                                       & =\dfrac{\dfrac{-2z\Delta z + o(\Delta z)}{(z^2-1)[(z+\Delta z)^2-1]}}{\Delta z} \\
		                                       & =\dfrac{-2z}{(z^2-1)^2}
	\end{aligned}
$$
在$z\neq\pm 1$处可导.

(3)$\dfrac{az+b}{cz+d}(c \cdot d \neq 0)$
$$
	\begin{aligned}
		f' & =\dfrac{ac-bd}{(cz+d)^2}
	\end{aligned}
$$
仅在$-\dfrac{c}{d}$处不可导.

(4)
$\bar{z}$
$$
	\begin{aligned}
		\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta z} & =\dfrac{\overline{\Delta z}}{\Delta z}
	\end{aligned}
$$
当$z+z_0$沿着平行实轴的方向趋近$z_0$时上式极限为$1$,而沿着虚轴趋近得到的极限为$-1$,从而处处不可导.

(5)
$|z|^2z$
当$\Delta z \rightarrow 0 $时:
$$
	\begin{aligned}
		\dfrac{f(z_0+\Delta z)-f(z_0)}{\Delta z} & =\dfrac{(z_0+\Delta z)^2(\overline{z_0+\Delta z})-z_0^2\overline{z_0}}{\Delta z}                                                     \\
		                                         & =\dfrac{2\Delta z z_0\bar{z_0}+\Delta z^2\bar{z_0}+\bar{\Delta z}z_0^2+2\Delta z \bar{\Delta z}z_0+\Delta z\bar{\Delta z}}{\Delta z} \\
		                                         & =2|z|^2+\dfrac{\bar{\Delta z}}{\Delta z}z_0^2
	\end{aligned}
$$
由(4)可知$\dfrac{\bar{\Delta z}}{\Delta z}$的极限并不存在,从而当且仅当$z_0=0$时$f(z)$在$z_0$处导数存在.

\section{讨论下列函数在何处满足C-R方程?}
 (1)$3-z+2z^2$

\textbf{解$1^{\circ}$:}\quad
$w' = -1 + 4z$,在复平面内可导,从而在复平面上任意处$u,v$都可微,且满足C-R方程.

\textbf{解$2^{\circ}$:}\quad

$$
	\begin{aligned}
		w & = 3 - z + 2z^2                       \\
		  & =3 - x + 2 x^2 - 2 y ^ 2 - yi + 4xyi
	\end{aligned}
$$

其中$u = 3 - x + 2 x ^2 - 2 y ^2,v = -y + 4xy$
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x}  & = \dfrac{\partial v}{\partial y}  \\
		                                & =-1 + 4x                          \\
		\dfrac{\partial u}{\partial y } & = -\dfrac{\partial v}{\partial x} \\
		                                & =-4y
	\end{aligned}
$$
在复平面上任意处都满足C-R方程.

(2)$\dfrac{1}{z}$

\textbf{解:}\quad
$w = \dfrac{x - yi}{x^2 + y^2}$.显然仅在$0$点处不存在导数,在其它任意点都可导,从而$u,v$在除$0$之外的点都可导,并且满足C-R方程.在$0$点处$w$不存在,偏导数不存在,不满足C-R方程.


(3)$x$
\textbf{解:}\quad
$\dfrac{\partial u}{\partial x} =1,\dfrac{\partial v}{\partial y}=0$,从而处处不满足.


(4)$2x^3+3y^3i$
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x} & =6x^2                                \\
		\dfrac{\partial v}{\partial y} & =9y^2                                \\
		\dfrac{\partial u}{\partial y} & =\dfrac{\partial v}{\partial x} & =0
	\end{aligned}
$$
从而在$x=\pm\dfrac{\sqrt{6}}{2}$时满足C-R方程.

\section{如果函数 $f(z)$ 在区域 $D$ 内解析,且满足下列条件之一, 求证 $f(z)$ 在 $D$ 内必 为常数.}
 (1) $|f(z)|$ 在 $D$ 内是常数;

$f = u+vi,u^2+v^2=C$,从而对$x,y$求偏导得:
$$
	\begin{aligned}
		2u\cdot u_x + 2v\cdot v_x & =0 \\
		2u\cdot u_y + 2v\cdot v_y & =0
	\end{aligned}
$$
由于$f(z)$在区域$D$内解析,从而其在$D$内满足Cauchy-Riemann方程,将其代入上式得:
$$
	\begin{aligned}
		2u\cdot u_x - 2v\cdot u_y & =0 \\
		2u\cdot u_y + 2v\cdot u_x & =0
	\end{aligned}
$$
可得$f=0$或$u_x=u_y=v_x=v_y=0$,从而$f$是常数.

(2) $\overline{f(z)}$ 在 $D$ 内解析;

由于$f,\bar{f}$都在$D$内解析,从而有
$$
	\begin{aligned}
		\dfrac{\partial u}{\partial x} & = \dfrac{\partial v}{\partial y} = \dfrac{\partial -v}{\partial y}
	\end{aligned}
$$
从而可得上式中任意一项都为$0$,同理可得$\dfrac{\partial u}{\partial y},\dfrac{\partial v}{\partial x}$都为$0$,从而$f'=0(\mathrm{d}x+\mathrm{d}y)=0$,$f$为常值函数.

(3) $\operatorname{Re} f(z)$ 或 $\operatorname{Im} f(z)$ 在 $D$ 内是常数;
\begin{proof}
	两种情况类似,若实部为常数,则有:
	$$
		\begin{aligned}
			\dfrac{\partial u}{\partial x} & =\dfrac{\partial v}{\partial y}=0  \\
			\dfrac{\partial u}{\partial y} & =-\dfrac{\partial v}{\partial x}=0
		\end{aligned}
	$$
	从而$u_x,u_y,v_x,v_y=0$,从而$f'=0$,$f$在$D$上为常值函数.
\end{proof}



(4) $f(z)$ 的辐角在 $D$ 内是常数.
\begin{proof}
	若不能写成$u=kv$的形式,则可用第(3)题的结论证明.若$k=0$同理.

	令$u=kv$,则
	$$
		k\cdot v_x = v_y\\
		k\cdot v_y = - v_x
	$$
	由于$k\neq 0$,从而可得$u_x,u_y,v_x,v_y=0$,从而$f'=0$,$f$在$D$上为常值函数.
\end{proof}

\section{设函数 $f(z)$ 在区域 $D$ 内解析,试证
$$
	\left(\dfrac{\partial}{\partial x}|f(z)|\right)^{2}+\left(\dfrac{\partial}{\partial y}|f(z)|\right)^{2}=\left|f^{\prime}(z)\right|^{2}
$$}
\begin{proof}
	由于$f$在$D$内解析,从而$u_x=v_y,u_y=-v_x$,代入:
	$$
		\begin{aligned}
			\left(\dfrac{\partial}{\partial x}|f(z)|\right)^{2}+\left(\dfrac{\partial}{\partial y}|f(z)|\right)^{2} & =\dfrac{1}{u^2+v^2}\left[\left(u\cdot u_x+v\cdot v_x\right)^2+\left(u\cdot u_y+v\cdot v_y\right)^2\right] \\
			                                                                                                        & =\dfrac{(u^2+v^2)(u_x^2+v_x^2)+2uvu_xv_x+2uvu_yv_y}{u^2+v^2}                                              \\
			                                                                                                        & =u_x^2+v_x^2
		\end{aligned}
	$$
	而
	$$
		f'(z)=u_x+iv_x\\
		|f'(z)|^2=u_x^2+v_x^2
	$$
	从而
	$$
		\left(\dfrac{\partial}{\partial x}|f(z)|\right)^{2}+\left(\dfrac{\partial}{\partial y}|f(z)|\right)^{2}=\left|f^{\prime}(z)\right|^{2}
	$$
\end{proof}
\section{设 $f(z)=u(r, \theta)+\mathrm{i} v(r, \theta), z=r \mathrm{e}^{i \theta}$, 则函数 $f(z)$ 在 $z$ 可导的充分必要条件 是 $u(r, \theta), v(r, \theta)$ 在 $(r, \theta)$ 可微,且满足极坐标下的 $\mathrm{C}-\mathrm{R}$ 方程
  $$
	  \dfrac{\partial u}{\partial r}=\dfrac{1}{r} \dfrac{\partial v}{\partial \theta}, \quad \dfrac{\partial v}{\partial r}=-\dfrac{1}{r} \dfrac{\partial u}{\partial \theta}(r>0)
  $$
  并且有
  $$
	  f^{\prime}(z)=(\cos \theta-\mathrm{i} \sin \theta)\left(\dfrac{\partial u}{\partial r}+\mathrm{i} \dfrac{\partial v}{\partial r}\right)=\dfrac{r}{z}\left(\dfrac{\partial u}{\partial r}+\mathrm{i} \dfrac{\partial v}{\partial r}\right)
  $$}
\begin{proof}
	必要性:

	$f(z) = u(z)+iv(z) = u(r,\theta) + iv(r,\theta)$.在$z$处$f(z)$可导,则若令$z=x+iy$,则$f$对$x,y$存在偏导数,并且满足
	\begin{equation}
		\dfrac{\partial u}{\partial x}=\dfrac{\partial v}{\partial y}\qquad \dfrac{\partial u}{\partial y} = -\dfrac{\partial v}{\partial x}
	\end{equation}
	由于$x=r\cos\theta,y=r\sin\theta$,从而$f$对$r,\theta$也存在偏导数.下面分别沿着$r$轴和$\theta$的方向趋近$z$,则这两种趋近方式下求得的$f$的导数应相同.对于沿着$r$轴趋近的:
	$$
		\begin{aligned}
			f'(z) & =\lim\limits_{\Delta z\rightarrow 0}\dfrac{f(z+\Delta z)-f(z)}{\Delta z}                                                                                                                                                                                                                                                           \\
			      & =\lim\limits_{\Delta r\rightarrow 0}\dfrac{\Delta u+i\Delta v}{\Delta r\mathrm{e}^{i\theta}}                                                                                                                                                                                                                                       \\
			      & =\dfrac{1}{\mathrm{e}^{i\theta}}\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)
		\end{aligned}
	$$
	对于沿着$\theta$轴趋近的:
	$$
		\begin{aligned}
            f'(z)&=\dfrac{f(z+\Delta z)-f(z)}{\Delta z}\\
            &=\dfrac{\Delta u+i\Delta v}{ir\mathrm{e}^{i\theta}\Delta \theta}\\
			&=\dfrac{1}{ir\mathrm{e}^{i\theta}}\left(\dfrac{\partial u}{\partial \theta}+i\dfrac{\partial v}{\partial \theta}\right)
		\end{aligned}
	$$
	可得:
	$$
		\dfrac{\partial u}{\partial r}=\dfrac{1}{r}\dfrac{\partial v}{\partial \theta}\qquad \dfrac{1}{r}\dfrac{\partial u}{\partial \theta}=-\dfrac{\partial v}{\partial r}
	$$
	即得到了应该满足的C-R方程的形式.

	我们取$\Delta z$沿着$r$轴趋近$0$,即$\Delta z=\Delta r(\cos\theta+i\sin\theta)$,则有:
	$$
		f'(z)=\dfrac{1}{\mathrm{e}^{i\theta}}\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)=(\cos \theta-\mathrm{i} \sin \theta)\left(\dfrac{\partial u}{\partial r}+\mathrm{i} \dfrac{\partial v}{\partial r}\right)=\dfrac{r}{z}\left(\dfrac{\partial u}{\partial r}+\mathrm{i} \dfrac{\partial v}{\partial r}\right)
	$$

	充分性:

	当$u,v$在该点对$r,\theta$可微的时候,且满足极坐标下的C-R方程时,则有:
    $$
        \begin{aligned}
            \Delta z &= \mathrm{e}^{i\theta}\Delta r+ ir\mathrm{e}^{i\theta}\Delta \theta\\
            \Delta w &= \Delta u+i\Delta v+\alpha(|\Delta z|)\\
            &=\left(\dfrac{\partial u}{\partial r}\Delta r+\dfrac{\partial u}{\partial \theta}\Delta\theta\right)+i\left(\dfrac{\partial v}{\partial r}\Delta r+\dfrac{\partial v}{\partial \theta}\Delta \theta\right)+\alpha(|\Delta z|)\\
            &=\left(\dfrac{\partial u}{\partial r}-\dfrac{i}{r}\dfrac{\partial u}{\partial\theta}\right)\Delta r+\left(\dfrac{\partial u}{\partial \theta}+ir\dfrac{\partial u}{\partial r}\right)\Delta\theta+\alpha(|\Delta z|)\\
            &=\left(\dfrac{\partial u}{\partial r}-\dfrac{i}{r}\dfrac{\partial u}{\partial\theta}\right)(\Delta r + ir\Delta\theta)+\alpha(|\Delta z|)\\
            &=\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)(\Delta r + ir\Delta\theta)+\alpha(|\Delta z|)
        \end{aligned}
    $$
    从而
    $$
    \begin{aligned}
        f'(z)&=\lim\limits_{\Delta z\rightarrow 0}\dfrac{\Delta u+i\Delta v}{\Delta z}\\
        &=\dfrac{1}{\mathrm{e}^{ir}}\left(\dfrac{\partial u}{\partial r}+i\dfrac{\partial v}{\partial r}\right)
    \end{aligned}
    $$
    由定义可知$f$在$z$点可导.
\end{proof}
\textcolor{blue}{\begin{remark}
    有关必要性还可以通过平面直角坐标系的C-R方程的条件代入到$\dfrac{\partial u}{\partial r}$等的表达式中一样可以解出.需要注意的是我们研究的是$f(z)$关于某条轴(如$\theta$或$r$轴)的导数,但其本质上是利用了$\Delta z = \dfrac{\partial z}{\partial r}\Delta r$或$\Delta z= \dfrac{\partial z}{\partial \theta}\Delta \theta$,直接让$f$对$r$或$\theta$取偏导需要用$f'(z)=\lim\limits_{\Delta z\rightarrow 0}\dfrac{\Delta u+i\Delta v}{\Delta z}=\lim\limits_{\Delta r\rightarrow 0}\dfrac{\Delta u+i\Delta v}{\dfrac{\partial z}{\partial r}\Delta r}$来求,不能想当然的就将$\dfrac{\partial z}{\partial r}$视为$1$,这仅在平面直角坐标系中会有这么好的性质.
\end{remark}}


\section{若函数 $f(z), g(z)$ 在 $z_{0}$ 解析, 且 $f\left(z_{0}\right)=g\left(z_{0}\right)=0$, 但 $g^{\prime}\left(z_{0}\right) \neq 0$, 则
  $$
	  \lim _{z \rightarrow z_{0}} \dfrac{f(z)}{g(z)}=\dfrac{f^{\prime}\left(z_{0}\right)}{g^{\prime}\left(z_{0}\right)}
  $$}
\begin{proof}
	由于$f,g$在$z_0$处解析并且$g'(z_0)\neq 0$,从而:
	$$
		\begin{aligned}
			\lim\limits_{z\rightarrow z_0}\dfrac{f(z)}{g(z)} & =\lim\limits_{z\rightarrow z_0}\dfrac{\dfrac{f(z)-f(z_0)}{z-z_0}}{\dfrac{f(z)-f(z_0)}{z-z_0}} \\
			                                                 & =\dfrac{f'(z)}{g'(z)}
		\end{aligned}
	$$
\end{proof}


\section{设函数 $f(z)=m y^{3}+n x^{2} y+i\left(x^{3}+l x y^{2}\right)$ 是全平面上的解析函数, 试求 $l$, $m, n$ 的值.}
\textbf{解:}\quad
由于$f$在平面上解析,从而可利用C-R方程:
$$
\begin{aligned}
    \dfrac{\partial u}{\partial x}&=\dfrac{\partial v}{\partial y}\\
    \dfrac{\partial u}{\partial y}&=-\dfrac{\partial v}{\partial x}\\
    2nxy&=2lxy\\
    3my^2+nx^2&=-(3x^2+ly^2)
\end{aligned}
$$
解得$(l,m,n)=(-3,1,-3)$

\section{试证}
(1) $\sinh \left(z_{1}+z_{2}\right)=\sinh z_{1} \cosh z_{2}+\cosh z_{1} \sinh z_{2}$;
\begin{proof}
    $$
		\begin{aligned}
			\sinh z_1\cosh z_2+\cosh z_1\sinh z_2&=\dfrac{2\mathrm{e}^{z_1+z_2}-2\mathrm{e}^{-z_1-z_2}}{4}\\
			&=\dfrac{\mathrm{e}^{z_1+z_2}-\mathrm{e}^{-z_1-z_2}}{2}\\
			&=\sinh (z_1+z_2)
		\end{aligned}
	$$
\end{proof}

(2) $\cosh \left(z_{1}+z_{2}\right)=\cosh z_{1} \cosh z_{2}+\sinh z_{1} \sinh z_{2}$;
\begin{proof}
    $$
		\begin{aligned}
			\cosh z_1\cosh z_2 + \sinh z_1 \sinh z_2
			&=\dfrac{(\mathrm{e}^{z_1}+\mathrm{e}^{-z_1})\cdot(\mathrm{e}^{z_2}+\mathrm{e}^{-z_2})}{4}+\dfrac{(\mathrm{e}^{z_1}-\mathrm{e}^{-z_1})\cdot(\mathrm{e}^{z_2}-\mathrm{e}^{-z_2})}{4}\\
			&=\dfrac{\mathrm{e}^{z_1+z_2}+\mathrm{e}^{-z_1-z_2}}{2}\\
			&=\cosh(z_1+z_2)
		\end{aligned}
	$$
\end{proof}
(3) $\cosh ^{2} z-\sinh ^{2} z=1$;
\begin{proof}
    $$
		\begin{aligned}
			\dfrac{(\mathrm{e}^{z}+\mathrm{e}^{-z})^2}{4}-\dfrac{(\mathrm{e}^z-\mathrm{e}^{-z})}{4}=1
		\end{aligned}
	$$
\end{proof}
(4) $\operatorname{sech}^{2} z+\tanh ^{2} z=1$.
\begin{proof}
    $$
		\begin{aligned}
			\operatorname{sech}^2 z + \tanh^2 z &=\dfrac{1}{\cosh ^2 z}+\dfrac{\sinh^2 z
			}{\cosh^2 z}\\
			&=1
		\end{aligned}
	$$
\end{proof}
\section{试证}
(1) $\sinh z=-i\sin iz$ ;

(2) $\cosh z=\cos iz$ ;

(3) $\sin z=-i\sinh iz$ ;

(4) $\cos z=\cosh iz$.

在复分析中,定义$\mathrm{e}^{z}=\sum\limits_{n=0}^{\infty}\dfrac{z^n}{n!}$.可以证明$\mathrm{e}^{z}$在复平面上是全纯的.从而类似,形式地定义$\sin,\cos$,然后可以使用$\mathrm{e}^{z}$的级数进行表示.
$$
\cos z=\sum_{n=0}^{\infty}(-1)^{n} \frac{z^{2 n}}{(2 n) !}, \quad \text { and } \quad \sin z=\sum_{n=0}^{\infty}(-1)^{n} \frac{z^{2 n+1}}{(2 n+1) !}
$$
它们从而可写成这种形式:
$$
\cos z=\frac{e^{i z}+e^{-i z}}{2} \quad \text { and } \quad \sin z=\frac{e^{i z}-e^{-i z}}{2 i}
$$
这个公式也被称为\textbf{Euler formulas}.

使用以上结论不难得到几个问题的证明.这里不再赘述.

\section{证明}

(1) $(\sinh z)^{\prime}=\cosh z$;

(2) $(\cosh z)^{\prime}=\sinh z$.

同上.

\section{求下列函数的值}
(1) $\mathrm{e}^{3+\mathrm{i}}$;

\textbf{解:}\quad
$$
\begin{aligned}
	\mathrm{e}^{3+\mathrm{i}}&=\mathrm{e}^3(\cos 1+i\sin 1)
\end{aligned}
$$

(2) $\operatorname{Ln}(-3+4 i)$;

\textbf{解:}\quad
$$
\begin{aligned}
	\operatorname{Ln}(-3+4 i)&=\ln|z|+i\mathrm{Arg}z\\
	&=\ln 5-i\arctan \dfrac{4}{3}+(2k+1)\pi i
\end{aligned}
$$

(3) $\sin \mathrm{i}$;

\textbf{解:}\quad
$$
\begin{aligned}
	\sin \mathrm{i}&=\dfrac{\mathrm{e}^{-1}-\mathrm{e}^{1}}{2i}\\
	&=\dfrac{i(\mathrm{e}-\mathrm{e}^{-1})}{2}
\end{aligned}
$$

(4) $\cos (1+\mathrm{i})$;

\textbf{解:}\quad
$$
\begin{aligned}
	\cos (1+\mathrm{i})&=\dfrac{\mathrm{e}^{i-1}+\mathrm{e}^{1-i}}{2}\\
	&=\dfrac{1}{2}\left[\cos 1(\mathrm{e}+\mathrm{e}^{-1})+i\sin 1(\mathrm{e}^{-1}-\mathrm{e})\right]
\end{aligned}
$$

(5) $\mathrm{i}^{1+\mathrm{i}}$;

\textbf{解:}\quad
$$
\begin{aligned}
	i^{1+i}&=\mathrm{e}^{(1+i)\mathrm{Ln}i}\\
	&=\mathrm{e}^{-\frac{(4k+1)\pi}{2}}\cdot\mathrm{e}^{i\frac{(4k+1)\pi}{2}}
\end{aligned}
$$

其中$k\in \mathbb{Z}$

(6) $(1+i)^{i}$.

\textbf{解:}\quad
$$
\begin{aligned}
	(1+i)^{i}&=\mathrm{e}^{i\mathrm{Ln}(1+i)}\\
	&=\mathrm{e}^{i\left(\frac{1}{2}\ln 2+i\left(\frac{\pi}{4}+2k\pi\right)\right)}\\
	&=\mathrm{e}^{-\frac{\pi}{4}+2k\pi}\cdot\mathrm{e}^{\frac{1}{2}i\ln 2}
\end{aligned}
$$

\section{解下列方程:}
(1) $\mathrm{e}^{z}=1+\sqrt{3} \mathrm{i}$;

\textbf{解:}\quad
$$
\begin{aligned}
	z&=\mathrm{Ln}(1+\sqrt{3}i)\\
	&=\ln 2 + i\left(\dfrac{\pi}{3}+2k\pi\right)
\end{aligned}
$$


(2) $\ln z=\dfrac{\pi}{2} \mathrm{i}$.

\textbf{解:}\quad
$$
\begin{aligned}
	z & = \mathrm{e}^{i\frac{\pi}{2}}\\
	&=i
\end{aligned}
$$
\end{document}